SATHEE: Ellipse 2 Question 8

Ellipse 2 Question 8

8. The eccentricity of an ellipse whose centre is at the origin is $1 / 2$ . If one of its directrices is $x = - 4$ , then the equation of the normal to it at $1, \frac{3}{2}$ is

(a) $2 y - x = 2$

(b) $4 x - 2 y = 1$

(d) $x + 2 y = 4$

(2017 Main)

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Solution:

We have, $e = \frac{1}{2}$ and $\frac{a}{e} = 4$

$∴ a = 2$

Now, $b^{2} = a^{2} (1 - e^{2}) = (2)^{2} 1 - {\frac{1}{2}}^{2} = 41 - \frac{1}{4} = 3$

$\Rightarrow b = \sqrt{3}$

$∴$ Equation of the ellipse is $\frac{x^{2}}{(2)^{2}} + \frac{y^{2}}{(\sqrt{3})^{2}} = 1$

$\Rightarrow \frac{x^{2}}{4} + \frac{y^{2}}{3} = 1$

Now, the equation of normal at $1, \frac{3}{2}$ is

$\begin{aligned} \frac{a^{2} x}{x_{1}} - \frac{b^{2} y}{y_{1}} & = a^{2} - b^{2} \\ \Rightarrow & \frac{4 x}{1} - \frac{3 y}{(3 / 2)} & = 4 - 3 \\ \Rightarrow & 4 x - 2 y & = 1 \end{aligned}$

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Ellipse 2 Question 8

8. The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x=−4, then the equation of the normal to it at 1,32 is

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8. The eccentricity of an ellipse whose centre is at the origin is $1 / 2$ . If one of its directrices is $x = - 4$ , then the equation of the normal to it at $1, \frac{3}{2}$ is